In this reaction , `(CH_(3))_(3) C - CH_(2)OH overset("Conc." H_(2)SO_(4))underset(170^(@)C)(to) X`
X is

Options

A. `(CH_3)_2C=CHCH_3`
B. `CH_3-=CH`
C. `(CH_3)_2CHCH_2CH_3`
D. `CH_3-CH_2-underset(CH_3)underset(|)C=CH_2`

Answer

To solve the problem, we need to analyze the reaction of the given tertiary alcohol, `(CH₃)₃C-CH₂OH`, when treated with concentrated sulfuric acid (H₂SO₄) at a temperature of 170°C. The goal is to determine the product X formed in this reaction.

### Step-by-Step Solution:

1. **Identify the Reactant:**
The reactant is `(CH₃)₃C-CH₂OH`, which is a tertiary alcohol. The structure can be represented as:
```
CH₃
|
CH₃-C-CH₂-OH
|
CH₃
```

2. **Understand the Role of Concentrated H₂SO₄:**
Concentrated sulfuric acid acts as a dehydrating agent. It removes water (H₂O) from the alcohol, leading to the formation of a carbocation intermediate.

3. **Formation of the Carbocation:**
When the alcohol is treated with H₂SO₄, the hydroxyl group (OH) is protonated, making it a better leaving group. The reaction proceeds as follows:
- Protonation of the OH group leads to the formation of an oxonium ion (OH₂⁺).
- The loss of water (H₂O) from the oxonium ion generates a carbocation.

4. **Determine the Stability of the Carbocation:**
The initial carbocation formed would be a primary carbocation, which is less stable. To achieve a more stable carbocation, a methyl shift occurs:
- A methyl group from the adjacent carbon shifts to the positively charged carbon, resulting in a more stable tertiary carbocation.

5. **Final Product Formation:**
The stable tertiary carbocation can then lose a proton (H⁺) to form an alkene. The elimination of H⁺ leads to the formation of the double bond:
- The resulting alkene structure will be:
```
CH₃
|
CH₃-C=CH-CH₃
|
CH₃
```
This can be represented as 2-methylbut-2-ene.

6. **Conclusion:**
The product X formed from the reaction of `(CH₃)₃C-CH₂OH` with concentrated H₂SO₄ at 170°C is 2-methylbut-2-ene.

### Final Answer:
**X is 2-methylbut-2-ene.**